Answer
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Hint: Rydberg equation is a mathematical relationship used to determine the wavelength of light emitted when an electron makes a transition form one energy level to another energy level. A photon of light is emitted when an electron moves from a higher energy level to lower energy level. Rydberg formula can be applied to the spectra of different elements.
The Rydberg’s constant is \[109677.57{\rm{ c}}{{\rm{m}}^{ - 1}}\].
Complete answer:
For a transition \[{n_2} \to {n_1}\] for an atom having atomic number Z, the wavelength is given by the following expression.
\[\dfrac{1}{\lambda } = R{Z^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]\]
Here R is Rydberg constant and the above equation is called Rydberg equation.
The atomic number of hydrogen is 1 and that of helium is 2.
Consider \[{n_2} \to {n_1}\] transitions in the hydrogen spectrum. The Rydberg equation will be
\[\dfrac{1}{\lambda } = R{\left( 1 \right)^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]\]
\[\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]\] … …(1)
Consider \[{n_4} \to {n_2}\] transitions in the hydrogen spectrum. The Rydberg equation will be
\[\dfrac{1}{\lambda } = R{\left( 2 \right)^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right]\]
\[\dfrac{1}{\lambda } = 4R\left[ {\dfrac{1}{4} - \dfrac{1}{{16}}} \right]\]
\[\dfrac{1}{\lambda } = R\left[ {1 - \dfrac{1}{4}} \right]\]
\[\dfrac{1}{\lambda } = \dfrac{{3R}}{4}\] … …(2)
The transition in hydrogen spectrum has the same wavelength as Balmer transition \[n = 4\;{\rm{ to }}\;n = 2{\rm{\;of\;H}}{{\rm{e}}^ + }\] spectrum
Hence, equation (1) = equation (2)
$R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{{3R}}{4}$
$\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{3}{4}$
For the first line in the Lyman series, \[{n_2} \to {n_1}\] or \[2 \to 1\].
$\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{3}{4}$
$\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right] = \dfrac{3}{4}$
$\left[ {1 - \dfrac{1}{4}} \right] = \dfrac{3}{4}$
$\dfrac{3}{4} = \dfrac{3}{4}$
Since both sides of the equation are equal, the transition is indeed the first line \[2 \to 1\] in the Lyman series of the hydrogen spectrum.
Hence, the transition of Lyman series in hydrogen spectrum have the same wavelength as Balmer transition \[n = 4\;{\rm{\; to }}\;n = 2\;{\rm{ of\;H}}{{\rm{e}}^ + }\] spectrum.
Note: In hydrogen spectrum, each transition is associated with a particular wavelength. Similarly, in the helium spectrum, each transition is associated with a particular wavelength. It is possible that the wavelength of a particular transition in the helium spectrum matches with the wavelength of a particular transition in the hydrogen atom. Properly chose the transition in the hydrogen atom that has the same wavelength as that of the given transition of the helium atom.
The Rydberg’s constant is \[109677.57{\rm{ c}}{{\rm{m}}^{ - 1}}\].
Complete answer:
For a transition \[{n_2} \to {n_1}\] for an atom having atomic number Z, the wavelength is given by the following expression.
\[\dfrac{1}{\lambda } = R{Z^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]\]
Here R is Rydberg constant and the above equation is called Rydberg equation.
The atomic number of hydrogen is 1 and that of helium is 2.
Consider \[{n_2} \to {n_1}\] transitions in the hydrogen spectrum. The Rydberg equation will be
\[\dfrac{1}{\lambda } = R{\left( 1 \right)^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]\]
\[\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]\] … …(1)
Consider \[{n_4} \to {n_2}\] transitions in the hydrogen spectrum. The Rydberg equation will be
\[\dfrac{1}{\lambda } = R{\left( 2 \right)^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right]\]
\[\dfrac{1}{\lambda } = 4R\left[ {\dfrac{1}{4} - \dfrac{1}{{16}}} \right]\]
\[\dfrac{1}{\lambda } = R\left[ {1 - \dfrac{1}{4}} \right]\]
\[\dfrac{1}{\lambda } = \dfrac{{3R}}{4}\] … …(2)
The transition in hydrogen spectrum has the same wavelength as Balmer transition \[n = 4\;{\rm{ to }}\;n = 2{\rm{\;of\;H}}{{\rm{e}}^ + }\] spectrum
Hence, equation (1) = equation (2)
$R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{{3R}}{4}$
$\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{3}{4}$
For the first line in the Lyman series, \[{n_2} \to {n_1}\] or \[2 \to 1\].
$\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{3}{4}$
$\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right] = \dfrac{3}{4}$
$\left[ {1 - \dfrac{1}{4}} \right] = \dfrac{3}{4}$
$\dfrac{3}{4} = \dfrac{3}{4}$
Since both sides of the equation are equal, the transition is indeed the first line \[2 \to 1\] in the Lyman series of the hydrogen spectrum.
Hence, the transition of Lyman series in hydrogen spectrum have the same wavelength as Balmer transition \[n = 4\;{\rm{\; to }}\;n = 2\;{\rm{ of\;H}}{{\rm{e}}^ + }\] spectrum.
Note: In hydrogen spectrum, each transition is associated with a particular wavelength. Similarly, in the helium spectrum, each transition is associated with a particular wavelength. It is possible that the wavelength of a particular transition in the helium spectrum matches with the wavelength of a particular transition in the hydrogen atom. Properly chose the transition in the hydrogen atom that has the same wavelength as that of the given transition of the helium atom.
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